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lys-0071 [83]
3 years ago
14

the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?

Chemistry
1 answer:
guapka [62]3 years ago
3 0

Answer : The solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

Explanation :

The solubility equilibrium reaction will be:

Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^{+}]^3[PO_4^{3-}]

K_{sp}=(3s)^3\times (s)

K_{sp}=27s^4

Given:

K_{sp} = 2.8\times 10^{-18}

Now put all the given values in the above expression, we get:

K_{sp}=27s^4

2.8\times 10^{-18}=27s^4

s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L

Therefore, the solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

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